Let's go through each part of the problem step-by-step: --- ### **(a) State the state of R-134a entering the evaporator** **Given:** - The expansion valve/capillary tube occludes flow, meaning **no work is done** at this stage - **State of R-a entering evaporator:** Saturated at -40°C. **Explanation:** Since the refrigerant enters the evaporator as saturated vapor at -40°C, it means: - It is in **vapor** phase (not mixed or liquid). - The process in the expansion valve is **isenthalpic** (constant enthalpy), but it does not involve work or heat transfer, only a pressure drop. **State of R-134a entering the evaporator:** - **Saturated vapor at -40°C**. --- ### **(b) Coefficient of performance for cooling \(\beta\)** The coefficient of performance (COP) for a refrigerator is given by: \[ \beta = \frac{Q_L}{W} \] Where: - \(Q_L\) is the heat removed from the low-temperature reservoir (evaporator). - \(W\) is the work input to the compressor. **Step 1: Find enthalpies at key points** - **State 1 (entering evaporator):** Saturated vapor at -40°C; \(h_1 = h_{g@ -40^\circ C}\). - **State 2 (leaving evaporator, entering compressor):** Superheated vapor at same pressure and temperature, approximately same enthalpy \(h_2 \approx h_1\). - **State 3 (after compression):** Superheated vapor at high pressure (~140 kPa). - **State 4 (after condenser):** Saturated liquid at 50°C. **Step 2: Use refrigerant tables or data for R-134a** - At -40°C (saturated vapor): - \(h_{g@ -40^\circ C} \approx 263.\) kJ/kg. - At 140 kPa (after compression): - \(h_2 \approx h_{g@ -40^\circ C} \approx 263\) kJ/kg (since enthalpy change during isentropic compression is small). - At 50°C (saturated liquid): - \(h_f \approx 74\) kJ/kg. **Step 3: Calculate \(Q_L\) and \(W\)** \[ Q_L = h_1 - h_4 = 263 - 74 = 189\, \text{kJ/kg} \] \[ W = h_2 - h_1 \approx 263 - 263 = \, \text{kJ/kg} \] However, in reality, \(h_2 > h_1\) due to compression work, but for simplicity, and because the process is ideal, the work input is: \[ W = h_2 - h_1 \] The actual value of \(h_2\) can be obtained from the superheated vapor tables at the high pressure and temperature after compression, but assuming isentropic compression, the enthalpy increase is small. **Final calculation:** \[ \beta = \frac{Q_L}{W} \] Using typical idealized data, the approximate COP for cooling: \[ \boxed{ \beta \approx \frac{189}{W} } \] The actual value depends on the detailed enthalpy data, but generally, for R-134a, COPs are around **3 to 4** under typical conditions. --- ### **(c) Mass flux rate of R-134a** Given: - Cooling cycle cools 10 kg of air from 30°C to 25°C in 10 minutes. - Air's specific heat \(c_{p,air} \approx 1.005\, \text{kJ/kg·K}\). **Step 1: Calculate heat removed \(Q_L\)** \[ Q_L = m_{air} \times c_{p,air} \times \Delta T \] \[ Q_L = 10\, \text{kg} \times 1.005\, \text{kJ/kg·K} \times (30 - 25)^\circ C \] \[ Q_L = 10 \times 1.005 \times 5 = 50.25\, \text{kJ} \] **Step 2: Power required** \[ \text{Power} = \frac{Q_L}{\text{time in seconds}} = \frac{50.25\, \text{kJ}}{600\, \text{sec}} \approx .084\, \text{kW} \] **Step 3: Mass flux rate of R-134a** Using the ideal cycle: \[ \dot{m}_{refrigerant} = \frac{\text{Cooling load}}{Q_L\, \text{per kg of refrigerant}} \] From part (b), the heat removed per unit mass of refrigerant is approximately \(Q_L \approx 189\, \text{kJ/kg}\). Thus, \[ \text{Mass flow rate of R-134a} = \frac{Q_L}{Q_{per\,kg}} \approx \frac{50.25\, \text{kJ}}{189\, \text{kJ/kg}} \approx .266\, \text{kg} \] **Final answer:** The mass flux rate of R-134a needed: \[ \boxed{ \approx .266\, \text{kg} \text{ over 10 minutes} } \] or approximately **.0266 kg/min**. --- **Summary:** - (a) R-134a enters the evaporator as saturated vapor at -40°C. - (b) The COP for cooling is approximately 3 to 4. - (c) The refrigerant mass flow rate required is about .266 kg over 10 minutes.